Test: MCQs (One or More Correct Option): Laws of Motion | JEE Advanced - JEE MCQ

This is a problem based on constraint motion. The motion of mass M is constraint with the motion of P and Q. Let AN = x, NO = z. Then velocity of mass is

From ΔANO, using pythagorous theorem

∴ x² + z² = ℓ²

Here x is a constant

Differentiating the above

equation w.r.t to t

Since earth is an accelerated frame and hence, cannot be an inertial frame.
Note : Strictly speaking Earth is accelerated reference frame. Earth is treated as a reference frame for practical examples and Newton's laws are applicable to it only as a limiting case.

Since the body is moving in a circular path therefore it needs centripetal force 

Also, the tangential acceleration acting on the mass is g sin θ.

At A the horizontal speeds of both the masses is the same. The velocity of Q remains the same in horizontal as no force is acting in the horizontal direction. But in case of P as shown at any intermediate position, the horizontal velocity first increases (due to N sin θ), reaches a max value at O and then decreases. Thus it always remains greater than v. Therefore t_P < t_Q.

The forces are resolved as shown in the figure.
When θ = 45º, sinθ = cosθ

The block will remain stationary and the frictional force is zero.
When θ > 45º, sinθ > cosθ
Therefore a frictional force acts towards Q.

As the bead is moving in the circular path

By energy conservation, 

From (1) and (2)
[N = mg cos θ - m [2g - 2g cosθ]
N = mg cos θ - 2mg + 2mg cosθ
N = 3mg cos θ-2mg
⇒ N = mg (3 cos θ-2)

Clearly N is positive (acts radially outwards) when 

Similarly, N acts radially inwards if